 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1/7b/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Write the complex similarity matrix as

S = X + iY,

with

X, Y

real. From

C = S D S^{- 1}

we obtain

CS = S D

, and equating real and imaginary parts gives

CX = X D, C Y = Y D .

Therefore

C (X + t Y) = (X + t Y) D

for every real

t

Now

p (t) = det (X + t Y)

is a real polynomial and

p (i) = det (X + iY) = det S \neq = 0.

Thus some real

t

has

p (t) \neq = 0

. Taking

T = X + t Y

, we have a real invertible matrix satisfying

CT = T D

, or

C = T D T^{- 1} .

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